Mapping Class Group of Knot Complements and Their Heegaard Splitting
Using this algorithm and the ideas from this paper (page 3) I gather that I can present $S^3-K$ where $K$ is the figure eight knot as the gluing of two genus 5 handle bodies, $H_5$, i.e.
$$
S^3-K=H_5cup_h H_5
$$ where $h$ is an element of the mapping class group of $S^3-K$.
Two questions:
(1) Suppose I'm given an element $h$ in the mapping class group of $S^3-K$. How do I take $h$ and produce a Heegaard splitting? I mean, every $h$ should be a prescription for mapping the meridians of one handle body (not necessarily of genus 5) to the parallels of the other right?
(2) Every mapping class group has the identity as an element. Since an element of the mapping class group is a homeomorphism of handle bodies, does that mean every manifold can be presented as the identity map on two handle bodies of the same genus?
EDIT: Warning, I've learned enough now to realize my first question isn't entirely clear. I'll see if I can't fix it up.
general-topology knot-theory low-dimensional-topology
asked Aug 13 at 23:44
Bob
66949
add a comment |
Using this algorithm and the ideas from this paper (page 3) I gather that I can present $S^3-K$ where $K$ is the figure eight knot as the gluing of two genus 5 handle bodies, $H_5$, i.e.
$$
S^3-K=H_5cup_h H_5
$$ where $h$ is an element of the mapping class group of $S^3-K$.
Two questions:
(1) Suppose I'm given an element $h$ in the mapping class group of $S^3-K$. How do I take $h$ and produce a Heegaard splitting? I mean, every $h$ should be a prescription for mapping the meridians of one handle body (not necessarily of genus 5) to the parallels of the other right?
(2) Every mapping class group has the identity as an element. Since an element of the mapping class group is a homeomorphism of handle bodies, does that mean every manifold can be presented as the identity map on two handle bodies of the same genus?
EDIT: Warning, I've learned enough now to realize my first question isn't entirely clear. I'll see if I can't fix it up.
general-topology knot-theory low-dimensional-topology
asked Aug 13 at 23:44
Bob
66949
1
Answer to your second question : No, with identity map on genus $g$ handlebodies, you only get $#^g S^1 times S^2$, the connected sum of $g$ $S^1times S^2$'s.
– Henry Park
Nov 29 at 19:29
1
I don't understand your first question; Heegaard splitting has something to do with mapping class group of surfaces, not 3-manifolds. So I don't get what you mean by taking an element of the mapping class group of the knot complement.
– Henry Park
Nov 29 at 19:32
@HenryPark Thanks Henry. Ya, I'm struggling a bit to understand low dimensional topology but since I asked this question I've learned enough to agree with you in that my question needs to be formulated better.
– Bob
Nov 30 at 0:06
add a comment |
Using this algorithm and the ideas from this paper (page 3) I gather that I can present $S^3-K$ where $K$ is the figure eight knot as the gluing of two genus 5 handle bodies, $H_5$, i.e.
$$
S^3-K=H_5cup_h H_5
$$ where $h$ is an element of the mapping class group of $S^3-K$.
Two questions:
(1) Suppose I'm given an element $h$ in the mapping class group of $S^3-K$. How do I take $h$ and produce a Heegaard splitting? I mean, every $h$ should be a prescription for mapping the meridians of one handle body (not necessarily of genus 5) to the parallels of the other right?
(2) Every mapping class group has the identity as an element. Since an element of the mapping class group is a homeomorphism of handle bodies, does that mean every manifold can be presented as the identity map on two handle bodies of the same genus?
EDIT: Warning, I've learned enough now to realize my first question isn't entirely clear. I'll see if I can't fix it up.
general-topology knot-theory low-dimensional-topology
asked Aug 13 at 23:44
Bob
66949
Using this algorithm and the ideas from this paper (page 3) I gather that I can present $S^3-K$ where $K$ is the figure eight knot as the gluing of two genus 5 handle bodies, $H_5$, i.e.
$$
S^3-K=H_5cup_h H_5
$$ where $h$ is an element of the mapping class group of $S^3-K$.
Two questions:
(1) Suppose I'm given an element $h$ in the mapping class group of $S^3-K$. How do I take $h$ and produce a Heegaard splitting? I mean, every $h$ should be a prescription for mapping the meridians of one handle body (not necessarily of genus 5) to the parallels of the other right?
(2) Every mapping class group has the identity as an element. Since an element of the mapping class group is a homeomorphism of handle bodies, does that mean every manifold can be presented as the identity map on two handle bodies of the same genus?
EDIT: Warning, I've learned enough now to realize my first question isn't entirely clear. I'll see if I can't fix it up.
general-topology knot-theory low-dimensional-topology
general-topology knot-theory low-dimensional-topology
asked Aug 13 at 23:44
Bob
66949
asked Aug 13 at 23:44
Bob
66949
edited Nov 30 at 0:09
asked Aug 13 at 23:44
Bob
66949
asked Aug 13 at 23:44
Bob
66949
asked Aug 13 at 23:44
Bob
66949
66949
1
Answer to your second question : No, with identity map on genus $g$ handlebodies, you only get $#^g S^1 times S^2$, the connected sum of $g$ $S^1times S^2$'s.
– Henry Park
Nov 29 at 19:29
1
I don't understand your first question; Heegaard splitting has something to do with mapping class group of surfaces, not 3-manifolds. So I don't get what you mean by taking an element of the mapping class group of the knot complement.
– Henry Park
Nov 29 at 19:32
@HenryPark Thanks Henry. Ya, I'm struggling a bit to understand low dimensional topology but since I asked this question I've learned enough to agree with you in that my question needs to be formulated better.
– Bob
Nov 30 at 0:06
add a comment |
1
Answer to your second question : No, with identity map on genus $g$ handlebodies, you only get $#^g S^1 times S^2$, the connected sum of $g$ $S^1times S^2$'s.
– Henry Park
Nov 29 at 19:29
1
I don't understand your first question; Heegaard splitting has something to do with mapping class group of surfaces, not 3-manifolds. So I don't get what you mean by taking an element of the mapping class group of the knot complement.
– Henry Park
Nov 29 at 19:32
@HenryPark Thanks Henry. Ya, I'm struggling a bit to understand low dimensional topology but since I asked this question I've learned enough to agree with you in that my question needs to be formulated better.
– Bob
Nov 30 at 0:06
1
1
Answer to your second question : No, with identity map on genus $g$ handlebodies, you only get $#^g S^1 times S^2$, the connected sum of $g$ $S^1times S^2$'s.
– Henry Park
Nov 29 at 19:29
Answer to your second question : No, with identity map on genus $g$ handlebodies, you only get $#^g S^1 times S^2$, the connected sum of $g$ $S^1times S^2$'s.
– Henry Park
Nov 29 at 19:29
1
1
I don't understand your first question; Heegaard splitting has something to do with mapping class group of surfaces, not 3-manifolds. So I don't get what you mean by taking an element of the mapping class group of the knot complement.
– Henry Park
Nov 29 at 19:32
I don't understand your first question; Heegaard splitting has something to do with mapping class group of surfaces, not 3-manifolds. So I don't get what you mean by taking an element of the mapping class group of the knot complement.
– Henry Park
Nov 29 at 19:32
@HenryPark Thanks Henry. Ya, I'm struggling a bit to understand low dimensional topology but since I asked this question I've learned enough to agree with you in that my question needs to be formulated better.
– Bob
Nov 30 at 0:06
@HenryPark Thanks Henry. Ya, I'm struggling a bit to understand low dimensional topology but since I asked this question I've learned enough to agree with you in that my question needs to be formulated better.
– Bob
Nov 30 at 0:06
add a comment |
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Answer to your second question : No, with identity map on genus $g$ handlebodies, you only get $#^g S^1 times S^2$, the connected sum of $g$ $S^1times S^2$'s.
– Henry Park
Nov 29 at 19:29
1
I don't understand your first question; Heegaard splitting has something to do with mapping class group of surfaces, not 3-manifolds. So I don't get what you mean by taking an element of the mapping class group of the knot complement.
– Henry Park
Nov 29 at 19:32
@HenryPark Thanks Henry. Ya, I'm struggling a bit to understand low dimensional topology but since I asked this question I've learned enough to agree with you in that my question needs to be formulated better.
– Bob
Nov 30 at 0:06